Question #01170
1 Answer
The
Explanation:
Because ammonium
Step 1: find
#K_"a" * K_"b" = 10^-14#
#K_"a"# of#("NH"_4)^(+)=10^-14/(K_text(b))=5.6 xx 10^-10#
Step 2: Set up the dissociation of ammonium equilibrium, and find the amount of
#{: (," ",bb("NH"_4^+),bb->,bb"NH"_3,+,bb("H"^+)), ("initial:",,"0.100 M",,"0 M",,"0 M"), ("change:",,-x,,+x,,+x), ("equilibrium:",,0.1-x,,x,,x) :}#
You can approximate that
#0.1 - x ~~ 0.1#
b/c
#K_"a" = (["NH"_3] * ["H"^+])/(["NH"_4^+]) = (x * x) /0.1 = 5.6 xx 10^-10#
#x^2 = 0.1 * 5.6 xx 10^-10#
#x = sqrt(5.6 xx 10^-11) =7.5 xx 10^-6#
#["H"^+] = x = 7.5 xx 10^-6#
Step 3: calculate
#"pH" = -log["H"^+] = -log(7.5xx10^-6) = 5.1#